Lyndal hits the target on average 2 out of every 3 shots in archery competitions - HSC - SSCE Mathematics Extension 1 - Question 4 - 2002 - Paper 1

To find the probability that Lyndal hits the target fewer than 9 times, we can calculate it using the complementary probability:

$P(X < 9) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8).$

Using the binomial probability formula for each value:

P(X < 9) = \sum_{k=0}^{8} {10 rack k} \left( \frac{2}{3} \right)^k \left( \frac{1}{3} \right)^{10-k}.

Leave this expression in unsimplified form.

By Vieta's formulas for the polynomial $P(x) = x^3 - 2x^2 + kx + 24$, the sum of the roots $\alpha + \beta + \gamma$ can be found as follows:

$\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{leading coefficient}} = -\frac{-2}{1} = 2.$

Again, using Vieta's formulas, product of the roots $\alpha\beta\gamma$ is:

$\alpha\beta\gamma = -\frac{\text{constant term}}{\text{leading coefficient}} = -\frac{24}{1} = -24.$

Let the equal roots be $r$ and $-r$ (since they are equal in magnitude but opposite in sign). Thus, using Vieta's:

1. The sum of the roots is: $r + (-r) + (\text{third root}) = 2.$ Therefore, the third root must be $2$.

2. We also know: $\alpha\beta\gamma = r(-r)(2) = -24$ Simplifying gives: $-2r^2 = -24\Rightarrow r^2 = 12 \Rightarrow r = 2\sqrt{3}.$

3. Finally substituting back, we can find $k$ using the equation: $\alpha + \beta + \gamma = 2 \Rightarrow k = -24 - r^2.$ Evaluating gives the specific value for $k$.

Starting from the differential equation for the particle: $\ddot{x} = -16x$ The characteristic equation leads us to a solution involving sinusoidal functions, specifically: $x(t) = A cos(4t) + B sin(4t)$ where $A$ and $B$ are determined by initial conditions. Given $x(0) = 1$ and rac{dx}{dt}|_{t=0} = 4, we can find $A$ and $B$.

After computing, substituting these values will yield the required form. The derived relationship will be $|x(t)| = 4\sqrt{2 - (A cos(4t) + B sin(4t))^2}$.

The greatest displacement occurs at the maximum of the function. Given the trigonometric nature, the maximum will occur when $sin$ or $cos$ takes the optimal value. We can compute: $\text{Maximum displacement} = 4\sqrt{2}.$

The general form for $x(t)$ will depend on the initial conditions and the derived solution: $x(t) = A\cos(4t) + B\sin(4t),$ where the constants $A$ and $B$ are derived from their respective initial displacements and velocities.